Find 10 Terms of the Laurent Expansion of 1/z(1-z)^2 in |z|>1

[bluepilotg-2_07]

Homework Statement

Expand #f(z) = \frac{1}{z(1-z)^2}# in a Laurent series over #-5 <= k <= 5# in the region #|z| > 1#

Relevant Equations

Laurent series

Since $|z| > 1$ we can rewrite as $|\frac{1}{z}|<1$. I rewrite the function as $$\frac{1}{z} \frac{1}{z^2(1-\frac{1}{z})^2}$$. For a typical geometric series: $$\frac{1}{1-\frac{1}{z}}=\sum_{k=0}\left(\frac{1}{z}\right)^k$$. The derivative is: $$\frac{-1}{z^2(1-\frac{1}{z})^2}=\sum_{k=1}-kz^{-k-1}$$.Therefore:$$\frac{1}{z}\frac{1}{z^2(1-\frac{1}{z})^2}=\sum_{k=1}k\frac{1}{z^{k+2}}$$. This implies the terms $-5 \leq k \leq 0$ should be zero and the remaining should be:$$z^{-3}+2z^{-4}+3z^{-5}+4z^{-6}+5z^{-7}$$. I do not understand why this is being marked wrong.

 

[renormalize]

I fixed the LaTeX of your homework statement:

Homework Statement: Expand $f(z) = \frac{1}{z(1-z)^2}$ in a Laurent series over $-5 <= k <= 5$ in the region $|z| > 1$

Laurent series are defined as expansions about a particular point $z=z_0\,$. For this problem, what is $z_0\text{:}$ $0\,$? $1\,$? Some other value?

 

[bluepilotg-2_07]

I fixed the LaTeX of your homework statement:

Laurent series are defined as expansions about a particular point $z=z_0\,$. For this problem, what is $z_0\text{:}$ $0\,$? $1\,$? Some other value?

I believe it is expanded about zero.

 

[renormalize]

I believe it is expanded about zero.

OK, then the 11-term sum asked for in the homework statement has the form:$$\sum_{k=-5}^{5}a_{k}z^{k}$$Do you have the right number of terms in your final expression of post #1?

 

[FactChecker]

As @renormalize points out, you may have too many terms in your answer. When you find that a term coefficient is zero, haven't you found that term and shouldn't you count it?

 

[bluepilotg-2_07]

OK, then the 11-term sum asked for in the homework statement has the form:$$\sum_{k=-5}^{5}a_{k}z^{k}$$Do you have the right number of terms in your final expression of post #1?

I believe so. My derivation implies terms from -5 to 0 are all zero, so everything I wrote are the non-zero terms.

 

[renormalize]

I believe so. My derivation implies terms from -5 to 0 are all zero, so everything I wrote are the non-zero terms.

How is that statement consistent with the expression in your first post:

the remaining should be:$$z^{-3}+2z^{-4}+3z^{-5}+4z^{-6}+5z^{-7}$$. I do not understand why this is being marked wrong.

That sum clearly has the powers $z^k$ for $-7\le k\le -3$; i.e., you have nothing but negative $k$ terms!
In other words: don't forget that, because $-5\le k\le 5$, you must write your 11-term sum in the specific form $\sum_{k=-5}^{5}a_{k}z^{k}$ to find the truncation of the Laurent series $\sum_{k=-\infty}^{\infty}a_{k}z^{k}$ that the problem is asking for.

 

[FactChecker]

If terms with zero coefficients do not count toward the 10, then the negative indices do not count. If they do count then there are too many terms with positive indices.